Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using std::min;
using std::vector;
template <class type>
inline const void read(type &in) {
  in = 0;
  char ch = getchar();
  bool f = 0;
  while (ch < 48 || ch > 57) {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch > 47 && ch < 58)
    in = (in << 3) + (in << 1) + (ch & 15), ch = getchar();
  if (f) in = -in;
}
template <class type>
inline const void write(type out) {
  if (out < 0) putchar('-'), out = -out;
  if (out > 9) write(out / 10);
  putchar(out % 10 + 48);
}
const int N = 1e5 + 10;
int n, m, q;
vector<int> citizen[N];
class Persistent_Segment_Tree {
 private:
  struct tree {
    int sum;
    tree *lson, *rson;
  } memory_pool[N * 20], *tail, *null, *root[N];
  inline const void init() {
    tail = memory_pool;
    null = tail++;
    root[0] = null->lson = null->rson = null;
  }
  inline const void update(tree *&now, tree *pre, int l, int r, int pos) {
    now = tail++;
    *now = *pre;
    now->sum++;
    if (l == r) return;
    int mid = l + r >> 1;
    if (pos <= mid)
      update(now->lson, pre->lson, l, mid, pos);
    else
      update(now->rson, pre->rson, mid + 1, r, pos);
  }
  inline const int query(tree *a, tree *b, tree *c, tree *d, int l, int r,
                         int k) {
    if (l == r) return l;
    int mid = l + r >> 1,
        lsum = a->lson->sum + b->lson->sum - c->lson->sum - d->lson->sum;
    if (k <= lsum)
      return query(a->lson, b->lson, c->lson, d->lson, l, mid, k);
    else
      return query(a->rson, b->rson, c->rson, d->rson, mid + 1, r, k - lsum);
  }

 public:
  inline Persistent_Segment_Tree() { init(); }
  inline const void update(int x, int y, int z) {
    update(root[x], root[y], 1, m, z);
  }
  inline const void follow(int x, int y) { root[x] = root[y]; }
  inline const int query(int a, int b, int lca, int falca, int k) {
    return query(root[a], root[b], root[lca], root[falca], 1, m, k);
  }
} pst;
int head[N], edc, next[N << 1], to[N << 1];
inline const void link(int u, int v) {
  next[++edc] = head[u];
  to[head[u] = edc] = v;
  next[++edc] = head[v];
  to[head[v] = edc] = u;
}
int fa[N], top[N], dep[N], sum[N], wson[N], size[N];
inline const void dfs(int p) {
  size[p] = 1;
  if (citizen[p].empty())
    pst.follow(p, fa[p]);
  else
    pst.update(p, fa[p], *citizen[p].begin());
  for (int i = 1; i < citizen[p].size(); i++) pst.update(p, p, citizen[p][i]);
  for (int i = head[p]; i; i = next[i]) {
    int son = to[i];
    if (son == fa[p]) continue;
    dep[son] = dep[p] + 1;
    fa[son] = p;
    sum[son] = sum[p] + citizen[son].size();
    dfs(son);
    size[p] += size[son];
    if (size[son] > size[wson[p]]) wson[p] = son;
  }
}
inline const void dfs(int p, int tp) {
  top[p] = tp;
  if (wson[p]) dfs(wson[p], tp);
  for (int son, i = head[p]; i; i = next[i])
    if (!top[son = to[i]]) dfs(son, son);
}
inline const int getlca(int a, int b) {
  while (top[a] ^ top[b])
    dep[top[a]] > dep[top[b]] ? a = fa[top[a]] : b = fa[top[b]];
  return dep[a] < dep[b] ? a : b;
}
inline const int pathkth(int a, int b, int lca, int k) {
  return pst.query(a, b, lca, fa[lca], k);
}
inline const void query(int a, int b, int k) {
  int lca = getlca(a, b);
  k = min(k, sum[a] + sum[b] - sum[lca] - sum[fa[lca]]);
  write(k);
  putchar(' ');
  for (int i = 1; i <= k; i++) write(pathkth(a, b, lca, i)), putchar(' ');
  putchar('\n');
}
int main() {
  read(n);
  read(m);
  read(q);
  for (int u, v, i = 1; i < n; i++) read(u), read(v), link(u, v);
  for (int c, i = 1; i <= m; i++) read(c), citizen[c].push_back(i);
  sum[1] = citizen[1].size();
  dfs(1);
  dfs(1, 1);
  for (int u, v, a; q--;) read(u), read(v), read(a), query(u, v, a);
  return 0;
}